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If $p\in\mathbb{Z}$ is a prime, let $\mathbb{Z}_p$ represents all rational numbers of the form $a/b$ where $p\nmid b$; $\mathbb{Z}_p$ is a ring. $a/b$ is a unit of $\mathbb{Z}_p$ if there is some element $c/d\in\mathbb{Z}_p$ so that $a/b\cdot c/d=1$.

I want to show if $a/b$ isn't a unit of $\mathbb{Z}_p$, then $a/b+1$ is a unit of $\mathbb{Z}_p$. This explains why Euclid's proof of the infinitude of primes breaks down in general for an integral domain.

This is a partial follow-up question from a previous question I submitted.

I tried to show this by proving the contrapositive:

To do this, I need to justify that if $a/b+1$ is not a unit of $\mathbb{Z}_p$, then $a/b$ is a unit.

If $a/b+1$ is not a unit, then $\nexists$ some $c/d\in\mathbb{Z}_p$ so that $(a/b+1)\cdot c/d=1$. I think I want to eventually use $a/b\cdot c/d=1\Rightarrow a/b+1=a/b+ac/bd\Rightarrow1=ac/bd=a/b\cdot c/d$, which would mean that $a/b$ is a unit.

I'm not totally sure about the breakdown for general integral domains. Obviously an integral domain is a nonzero commutative ring with no zero divisors. I think the issue in $\mathbb{Z}_p$ falls from that since we constructed a ring that essentially has "one" prime, all other primes are expressed as a unit times some power of p. If we do the Euclid trick of "add one and find a contradiction," then we end up with an element $a^*/b^*+1$ that isn't a unit in $\mathbb{Z}_p$. If this is the case, how would I be able to state this in general?

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    $\mathbb{Z}p$ is overloaded. $\mathbb{Z}{(p)}$ is a more standard notation for this ring; it's called the localisation at $(p)$. – lhf Jun 18 '21 at 17:37
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    $a/b,$ is a nonunit iff $,p\mid a,,$ so $,a/b+1 = (a!+!b)/b,$ is a unit $!\iff! p\nmid a!+!b,,$ so the claim is equivalent to $,p\mid a\Rightarrow p\nmid a!+!b,,$ true by $,p\nmid b\ \ $ – Bill Dubuque Jun 18 '21 at 17:46
  • That makes sense, thank you. –  Jun 18 '21 at 17:53
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    More generaly, see the Note here and the role it plays in the ideal theoretic generalization of Euclid's proof there. – Bill Dubuque Jun 18 '21 at 18:00
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    Following the idea in the theorem in the note, we can view the inference more conceptually as follows: $,a/b,$ nonunit $\Rightarrow a/b,$ is divisible by $,p,$ (i.e. by every prime) so $,1+a/b,$ is not divisible by any prime, so it is a unit. If you are familiar with ideals I recommend that you peruse the theorem and the generalization of Euclid's proof linked above. – Bill Dubuque Jun 18 '21 at 18:25

1 Answers1

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Let us go in a straightforward way.

Suppose $a/b$ is not a unit. I contend that $p$ must divide $a$. Indeed, if $p \nmid a$, then $b/a \in \Bbb Z_p$ and thus, $a/b$ has an inverse.
Now, since $p \mid a$ and $p \nmid b$, this means that $p \nmid (a + b)$. However, note that $$\frac{a}{b} + 1 = \frac{a + b}{b}.$$ Since $p \nmid (a + b)$, we see that $b/(a + b) \in \Bbb Z_p$ and thus $\frac{a}{b} + 1$ is a unit.

This breaks down the proof of Euclid because given primes $r_1, \ldots, r_n \in \Bbb Z_p$, the product $r_1 \cdots r_n$ is not a unit. In turn, $r_1 \cdots r_n + 1$ is a unit and thus, has no prime factor.

If you do wish to prove via contrapositive, note that in the above we implicitly made the following observation:

Given $a/b \in \Bbb Z_p$, it is a unit iff $p \nmid a$.

Edit: Bill Dubuque has mentioned this in their comment on the original question.

Aryaman Maithani
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    When you duplicate arguments already in comments you should mention / ack them – Bill Dubuque Jun 18 '21 at 17:55
  • Thanks for your explanation, that was very intuitive and clearly explained. My main issue is I overlooked the $p\nmid a$ step, which would make the next steps more apparent. –  Jun 18 '21 at 17:56
  • @BillDubuque: I actually hadn't read your comment. Have edited the post now since you'd posted the comment before I'd started writing my answer. Out of curiosity: 1. Why didn't you put your comment as an answer in itself? To move the question out of the unanswered queue, it would have to be answered anyway and it looks as if any answer would use similar arguments. 2. Since comments are, by nature, supposed to be temporary, it does not actually make much sense to quote a comment that has outlived its purpose. – Aryaman Maithani Jun 18 '21 at 18:03
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    Because comments are helpful in determining the knowledge level of the questioner, to help determine the best level of exposition. I had planned to do so but that opportunity was - alas - ruined by your FGITW duplicate answer (see my subsequent comments). Please strive to respect discussion in comments. – Bill Dubuque Jun 18 '21 at 18:16
  • So sorry to reopen the comments on this one. Why is $r_1\cdots r_n$ not a unit if all primes of this ring are $p$ times some unit $a/b$? Is it because $p$ divides whatever is the numerator of this product? –  Jun 21 '21 at 17:22
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    @DullandWitlessBoi: That's just a fact about rings in general. A unit is not divisible by any non-unit. In particular, if a prime $p$ divides some element $a$, then $a$ cannot be a unit. (Again, in general, if $b$ divides a unit, then $b$ must itself be a unit. (Why?) Equivalently, a non-unit cannot divide a unit. Since primes are non-units (by definition), it follows that a prime cannot divide a unit.) – Aryaman Maithani Jun 21 '21 at 21:42